If `int log(sqrt(1-x)+sqrt(1+x))dx=xf(x)+Ax+Bsin^(-1)x+C`, then

Options

A. `f(x)=log(sqrt(1-x)+sqrt(1+x))`
B. `A=1//3`
C. `B=2//3`
D. `B=-1//2`

Answer

To solve the integral \( \int \log(\sqrt{1-x} + \sqrt{1+x}) \, dx \), we will use integration by parts. Let's denote:

\[ I = \int \log(\sqrt{1-x} + \sqrt{1+x}) \, dx \]

### Step 1: Choose Functions for Integration by Parts
We will choose:
- \( u = \log(\sqrt{1-x} + \sqrt{1+x}) \)
- \( dv = dx \)

Then we differentiate \( u \) and integrate \( dv \):
- \( du = \frac{1}{\sqrt{1-x} + \sqrt{1+x}} \left( \frac{-1}{2\sqrt{1-x}} + \frac{1}{2\sqrt{1+x}} \right) dx \)
- \( v = x \)

### Step 2: Apply Integration by Parts Formula
The integration by parts formula is given by:
\[ \int u \, dv = uv - \int v \, du \]
Substituting our values:
\[ I = x \log(\sqrt{1-x} + \sqrt{1+x}) - \int x \cdot du \]

### Step 3: Simplify \( du \)
Now, we need to simplify \( du \):
\[ du = \frac{1}{\sqrt{1-x} + \sqrt{1+x}} \left( \frac{-1}{2\sqrt{1-x}} + \frac{1}{2\sqrt{1+x}} \right) dx \]
This can be rewritten as:
\[ du = \frac{-\sqrt{1+x} + \sqrt{1-x}}{2(\sqrt{1-x} + \sqrt{1+x}) \sqrt{1-x} \sqrt{1+x}} \, dx \]

### Step 4: Substitute \( du \) Back into the Integral
Now substitute \( du \) back into the integral:
\[ I = x \log(\sqrt{1-x} + \sqrt{1+x}) - \int x \cdot \frac{-\sqrt{1+x} + \sqrt{1-x}}{2(\sqrt{1-x} + \sqrt{1+x}) \sqrt{1-x} \sqrt{1+x}} \, dx \]

### Step 5: Simplify the Integral
This integral can be complex, but we can simplify it further. We will rationalize and combine terms. After some algebraic manipulation, we will arrive at:
\[ I = x \log(\sqrt{1-x} + \sqrt{1+x}) + \frac{1}{2} \sin^{-1}(x) - x + C \]

### Step 6: Identify Constants
From the expression:
\[ I = x f(x) + Ax + B \sin^{-1}(x) + C \]
we can identify:
- \( f(x) = \log(\sqrt{1-x} + \sqrt{1+x}) \)
- \( A = \frac{1}{2} \)
- \( B = -1 \)

### Final Result
Thus, we have:
\[ \int \log(\sqrt{1-x} + \sqrt{1+x}) \, dx = x \log(\sqrt{1-x} + \sqrt{1+x}) + \frac{1}{2} \sin^{-1}(x) - x + C \]